- By
**tieve** - Follow User

- 136 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Darcy meets theta' - tieve

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Darcy meets theta

### Briefly: how this is applied to gas integrand must be zero for all points

The plummeting fortunes of Permeability and other distress caused by inadequate fluids

Extending Darcy to Unsaturated Media

- 1907 Buckingham saw Darcy could describe unsaturated flow
- q = - K(q) ÑH [2.102]
- K(q): a function of the moisture content.
Conductivity is not a function of pressure: the geometry of the water filled pores is all that matters, which is dictated by q alone. To express K as a function of pressure you must employ the hysteretic functional relationship between q and H.

How does K vary with q?

- Drops like a rock.
- Three factors are responsible for this behavior:
- 1. Large pores empty first. These are the pores with least resistance to flow, since they have the largest diameters (recall the 1/r4 dependence of hydraulic resistance in the in the Hagen-Poiseuille equation).
- 2. Flow paths increase in length. Instead of proceeding straight through a chunk of media, the flow must avoid all the empty pores, making the path more tortuous.
- 3. There is less cross-section of flow. In any given area normal to flow, all the fluid must pass through a smaller portion of this area; for a given aerial flux the pore velocity must be higher.

Permeability with theta...

- To get a feel for how fast K drops, let’s consider the effect of tortuosity alone (item 2 above).
- Must be distinguished from the “Darcian” flow length and the “Darcian” velocity.

Difference between the true microscopic fluid flow path length and flow velocity in comparison with the “Darcian” values, which are based on a macroscopic picture of the system.

A few illustrative calculations…

- So writing the true pressure gradient that acts on the fluid we have
- Le/L ratio of the true path length to the Darcian length.
- Next write equation for capillary velocity, vf (Hagen-Poiseuille)
- Translating into the Darcian velocity, q, we find

Conclusions on tortuosity...

- Solving for q we find
- or, after comparing to Darcy’s law we see that
- where C is some constant.
- K goes down with (tortuosity)2 and (radius)2
- K also hit by big pores emptying first
- Conductivity will drop precipitously as q decreases

Characteristics of K(q)

1: at saturation K= Ks.

2-3: K(q) = 0: pendular water

Adding Conservation of Mass: Richards Equation Consider arbitrary volume media. Keep track of fluid going into and out of this volume.

- Need to add the constraint imposed by the conservation of mass (L.A. Richards 1931).
- There are many ways to obtain this result (we’ll do a couple).

- Volume is independent of time, bring the time derivative inside
- Apply the divergence theorem to the right side
- Combining [2.110] and [2.111]

- Since [2.112] is true for any volume element that the integrand must be zero for all points
- Replace q using Darcy’s law to obtain Richards equation
- where H is the total potential. This is also referred to as the Fokker-Plank equation

In terms of elevation and pressure integrand must be zero for all points

- Total potential is sum of gravity potential and pressure, H = h + z
- but
- So we obtain
- noticing that

And finally... integrand must be zero for all points

- So Richards Equation may be written (drum roll please....)
Before we can get anywhere we need the relationships, K(q) and h{q}.

- First order in time and second order in space; require
- 1 initial condition and
- 2 boundary conditions (top and bottom)

The diffusion form of R’s eq. integrand must be zero for all points

- Can put in more familiar form by introducing
- the soil diffusivity. Note that

The diffusion form of R’s eq. integrand must be zero for all points

- D(q) gives us a diffusion equation in q
- Favorite trick in solving diffusion problems is to assume that D is constant over space and pull it outside the derivative: it will find no place here!
- D(q) strongly non-linear function of q and varies drastically as a function of space.
- This makes Richards equation quite an interesting challenge in terms of finding tidy analytical solutions, and even makes numerical modelers wince a bit due to the very rapid changes in both D and q.

What about that no-slip boundary?

Gas Flow in Porous Media integrand must be zero for all points

- Might expect movement of gases is a simple extension of liquids
- use gas density & viscosity with intrinsic permeability to get K
- use Darcy's law.

- Well, it ain't quite that simple.
- Recall the "no-slip" boundary condition
- Idea: collisions between molecules in liquid so frequent that near a fixed surface the closest molecules will be constantly loosing all of their wall-parallel energy, rendering them motionless from the macroscopic perspective.
- Requisite: mean free path of travel short compared to the aperture through which the liquid is moving.

… about Gases integrand must be zero for all points

- Gases
- mean free length of travel on the order of media pore size.
- No-slip condition for liquids does not apply

- Taken together, results in "Klinkenberg effect" (experimental/theoretical 1941 paper by L.J. Klinkenberg).
- Gas permeability, Kg, a function of gas pressure (dictates mean free path length)

- k integrand must be zero for all points intrinsic permeability to liquid flow
- r characteristic radius apertures of the media
- l mean free path length of the gas molecules
- c proportionality factor between the mean free path for the free gas compared to the mean free path for gas which just collided with the capillary wall (c is just slightly less than 1, Klinkenberg, 1941).
- Coarse media: capillaries larger than the mean free path length, the intrinsic permeability for liquids is recovered
- Fine media: gas permeability exceeds liquid permeability.

Final notes on gas permeability integrand must be zero for all points

- kg/k measure of size of conducting pathways,
- can be used as a diagnostic parameter (Reeve, 1953).

- As PÞ0 the mean free path length lÞr
- limit for kg»10 k
- Klinkenberg's found max kg» 5 - 20 times k

- Practical purposes
- correction for vicinity of 1 bar 20 to 80% over the liquid permeability, depending upon the media (Klinkenberg, 1941).
- Methods of measurement of gas permeability:
- Corey (1986),
- New methods: Moore and Attenborough, 1992.
- kg depends on the liquid content, analogous to hydraulic conductivity with moisture content (see Corey, 1986).

Relative permeabilities of wetting and non-wetting phases integrand must be zero for all points

Ksnw

Ksw

Wetting Fluid Permeability

Non-Wetting Fluid Permeability

0

0

0

snw

snw

0

Expressions for Conductivity & Retention integrand must be zero for all points

- To solve Richards equation we need the mathematical relationships between pressure, moisture content and conductivity.
- Conductivity is a non-hysteretic function of moisture content
- Moisture content and pore pressure are related through a hysteretic functional (often hysteretic ignored; gives expressions of conductivity in terms of matric potential).
- For analytical solution, must use analytical expressions.
- For numerical solutions may use constructed as interpolations between successive laboratory data.

Download Presentation

Connecting to Server..